Digital Signal Processing Reference
In-Depth Information
Fig. 2.12 An example of a
recursive digital system
(leaky integrator)
x ( n )
y ( n )
z
−1
b
Example (2)) The difference equation that describes the recursive system in Fig.
(
2.12
) is given by:
y
ð
n
Þ¼
x
ð
n
Þþ
by
ð
n
1
Þ;
ð
2
:
14
Þ
where, b is a multiplicative constant (gain). Taking the ZT of both sides of (
2.14
)
yields:
Y
ð
z
Þ¼
X
ð
z
Þþ
bz
1
Y
ð
z
Þ:
Re-arranging terms leads to
½
1
bz
1
Y
ð
z
Þ¼
X
ð
z
Þ:
)H
ð
z
Þ¼
Y
ð
z
Þ
:
1
bz
1
¼
z
1
X
ð
z
Þ
¼
z
b
From Tables (z-Transform Pairs), the IZT is h(n) = b
n
u(n).
The system frequency response is:
H
ð
x
Þ¼
H
ð
z
Þj
z
¼
exp
ð
jxT
s
Þ
¼
e
jxT
s
e
jxT
s
b
¼
e
j2pfT
s
e
j2pfT
s
b
¼
e
j2pf
=
fs
e
j2pf
=
f
s
b
¼
e
j2pm
e
j2pm
b
¼
e
jX
e
jX
b
where m = f/f
s
and X
¼
2pm are the normalized cyclic frequency and normalized
angular frequency, respectively. The magnitude and phase responses are:
H
ð
x
Þj¼j
H
ð
e
jxT
s
Þj¼
j
e
jxT
s
j
1
j½
cos
ð
xT
s
Þ
b
j sin
ð
xT
s
Þj
j
e
jxT
s
b
j
¼
1
½
cos
ð
xT
s
Þ
b
2
½
sin
ð
xT
s
Þ
2
¼
q
\H
ð
x
Þ
b sin
ð
x
T
s
Þ
1
b cos
ð
x
T
s
Þ
¼
\H
ð
e
jxT
s
Þ¼
tan
1
As a special case, let b = 0.5 and f
s
= 1 Hz (i.e., T
s
= 1 s). At f = 0.1 Hz (i.e.,
x = 2p(0.1) = 0.6283rad/s), the magnitude response is
j
H
ð
e
jxT
s
Þj¼
1
:
506 and
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